3.15 \(\int (a+a \cos (c+d x))^2 (A+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=110 \[ \frac {a^2 (A+C) \tan (c+d x)}{d}+\frac {a^2 (A+2 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}+a^2 C x+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

[Out]

a^2*C*x+a^2*(A+2*C)*arctanh(sin(d*x+c))/d+a^2*(A+C)*tan(d*x+c)/d+1/3*A*(a^2+a^2*cos(d*x+c))*sec(d*x+c)*tan(d*x
+c)/d+1/3*A*(a+a*cos(d*x+c))^2*sec(d*x+c)^2*tan(d*x+c)/d

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Rubi [A]  time = 0.35, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3044, 2975, 2968, 3021, 2735, 3770} \[ \frac {a^2 (A+C) \tan (c+d x)}{d}+\frac {a^2 (A+2 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}+a^2 C x+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

a^2*C*x + (a^2*(A + 2*C)*ArcTanh[Sin[c + d*x]])/d + (a^2*(A + C)*Tan[c + d*x])/d + (A*(a^2 + a^2*Cos[c + d*x])
*Sec[c + d*x]*Tan[c + d*x])/(3*d) + (A*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 2975

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3044

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n +
2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b
*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0
])

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int (a+a \cos (c+d x))^2 (2 a A+3 a C \cos (c+d x)) \sec ^3(c+d x) \, dx}{3 a}\\ &=\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int (a+a \cos (c+d x)) \left (6 a^2 (A+C)+6 a^2 C \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{6 a}\\ &=\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int \left (6 a^3 (A+C)+\left (6 a^3 C+6 a^3 (A+C)\right ) \cos (c+d x)+6 a^3 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx}{6 a}\\ &=\frac {a^2 (A+C) \tan (c+d x)}{d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int \left (6 a^3 (A+2 C)+6 a^3 C \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=a^2 C x+\frac {a^2 (A+C) \tan (c+d x)}{d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\left (a^2 (A+2 C)\right ) \int \sec (c+d x) \, dx\\ &=a^2 C x+\frac {a^2 (A+2 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 (A+C) \tan (c+d x)}{d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [B]  time = 6.43, size = 748, normalized size = 6.80 \[ \frac {\sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^2 \left (5 A \sin \left (\frac {d x}{2}\right )+3 C \sin \left (\frac {d x}{2}\right )\right )}{12 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {\sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^2 \left (5 A \sin \left (\frac {d x}{2}\right )+3 C \sin \left (\frac {d x}{2}\right )\right )}{12 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {(-A-2 C) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^2 \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{4 d}+\frac {(A+2 C) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^2 \log \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{4 d}+\frac {\left (7 A \cos \left (\frac {c}{2}\right )-5 A \sin \left (\frac {c}{2}\right )\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^2}{48 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {\left (-5 A \sin \left (\frac {c}{2}\right )-7 A \cos \left (\frac {c}{2}\right )\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^2}{48 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {A \sin \left (\frac {d x}{2}\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^2}{24 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}+\frac {A \sin \left (\frac {d x}{2}\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^2}{24 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}+\frac {1}{4} C x \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^2 \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

(C*x*(a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4)/4 + ((-A - 2*C)*(a + a*Cos[c + d*x])^2*Log[Cos[c/2 + (d*x)/2
] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4)/(4*d) + ((A + 2*C)*(a + a*Cos[c + d*x])^2*Log[Cos[c/2 + (d*x)/2]
+ Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4)/(4*d) + (A*(a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*Sin[(d*x)/
2])/(24*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^3) + ((a + a*Cos[c + d*x])^2*Sec[c/2
+ (d*x)/2]^4*(7*A*Cos[c/2] - 5*A*Sin[c/2]))/(48*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)
/2])^2) + ((a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*(5*A*Sin[(d*x)/2] + 3*C*Sin[(d*x)/2]))/(12*d*(Cos[c/2]
- Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (A*(a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*Sin[(d
*x)/2])/(24*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^3) + ((a + a*Cos[c + d*x])^2*Sec
[c/2 + (d*x)/2]^4*(-7*A*Cos[c/2] - 5*A*Sin[c/2]))/(48*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 +
(d*x)/2])^2) + ((a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*(5*A*Sin[(d*x)/2] + 3*C*Sin[(d*x)/2]))/(12*d*(Cos[
c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

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fricas [A]  time = 2.11, size = 131, normalized size = 1.19 \[ \frac {6 \, C a^{2} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (5 \, A + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, A a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/6*(6*C*a^2*d*x*cos(d*x + c)^3 + 3*(A + 2*C)*a^2*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(A + 2*C)*a^2*cos(d
*x + c)^3*log(-sin(d*x + c) + 1) + 2*((5*A + 3*C)*a^2*cos(d*x + c)^2 + 3*A*a^2*cos(d*x + c) + A*a^2)*sin(d*x +
c))/(d*cos(d*x + c)^3)

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giac [A]  time = 0.53, size = 187, normalized size = 1.70 \[ \frac {3 \, {\left (d x + c\right )} C a^{2} + 3 \, {\left (A a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (A a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 8 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

1/3*(3*(d*x + c)*C*a^2 + 3*(A*a^2 + 2*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(A*a^2 + 2*C*a^2)*log(abs(
tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 8*A*a^2*tan(
1/2*d*x + 1/2*c)^3 - 6*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 9*A*a^2*tan(1/2*d*x + 1/2*c) + 3*C*a^2*tan(1/2*d*x + 1/2
*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

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maple [A]  time = 0.35, size = 134, normalized size = 1.22 \[ \frac {5 a^{2} A \tan \left (d x +c \right )}{3 d}+a^{2} C x +\frac {a^{2} C c}{d}+\frac {a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {a^{2} C \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)

[Out]

5/3*a^2*A*tan(d*x+c)/d+a^2*C*x+1/d*a^2*C*c+a^2*A*sec(d*x+c)*tan(d*x+c)/d+1/d*a^2*A*ln(sec(d*x+c)+tan(d*x+c))+2
/d*a^2*C*ln(sec(d*x+c)+tan(d*x+c))+1/3*a^2*A*sec(d*x+c)^2*tan(d*x+c)/d+1/d*a^2*C*tan(d*x+c)

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maxima [A]  time = 0.48, size = 138, normalized size = 1.25 \[ \frac {2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 6 \, {\left (d x + c\right )} C a^{2} - 3 \, A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A a^{2} \tan \left (d x + c\right ) + 6 \, C a^{2} \tan \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

1/6*(2*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 + 6*(d*x + c)*C*a^2 - 3*A*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 -
1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 6*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1))
+ 6*A*a^2*tan(d*x + c) + 6*C*a^2*tan(d*x + c))/d

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mupad [B]  time = 0.94, size = 184, normalized size = 1.67 \[ \frac {2\,A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {5\,A\,a^2\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^2)/cos(c + d*x)^4,x)

[Out]

(2*A*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d + (4*C*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (5*A*a^2*sin(c + d*x))/(3*d*cos(c + d*x)) +
(A*a^2*sin(c + d*x))/(d*cos(c + d*x)^2) + (A*a^2*sin(c + d*x))/(3*d*cos(c + d*x)^3) + (C*a^2*sin(c + d*x))/(d*
cos(c + d*x))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**2*(A+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

Timed out

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